Examples of other prime index

Caution

The statement is that if we have a subgroup whose index is the least prime divisor of the order of the group, that subgroup is normal. The statement does not say that among the subgroups of prime index, the one of least prime index is normal. For instance, in the alternating group on five letters, there is no subgroup of index two (the least prime divisor). There is also no subgroup of index three. There are subgroups of index five, namely A4 in A5, and these are not normal.

Proof

Proof using action on coset space

Given: A group and a subgroup such that , where is the least prime divisor of the order of .

To prove: is normal in .

Proof:

(Facts used: fact (1)): Consider the action of on the left coset space of , by left multiplication (Fact (1)). This gives a homomorphism where is the symmetric group on , which has size . The kernel of this homomorphism is a normal subgroup of contained inside (in fact, it is the normal core of ).

(Facts used: fact (2)):The image is a subgroup of , and hence, by fact (2), its order divides the order of . Thus, the order of divides .

(Facts used: fact (3)): The image is isomorphic to the quotient group , and thus, by fact (3), its order divides the order of . Thus, the order of divides the order of .

(Give data used: is the least prime divisor of the order of ): Since is the least prime divisor of the order of , we conclude that . Combining this with steps (2) and (3), we see that the order of divides . Since , we obtain that .

We thus have that , with and . This forces , yielding that is a normal subgroup of .

Proof using action on the set of conjugates

Now, since is a maximal subgroup in , is either normal or self-normalizing. Assume by contradiction that is not normal. Then it is self-normalizing. The same is true for .

Consider the set of all conjugates of in . Then, acts on by conjugation. Restricting to , acts on by conjugation.

Thus, every element of cannot normalize , and hence the action of on has no fixed points other than itself.

We further know that the total cardinality of is , and that there is exactly one fixed point. Thus, there is a nontrivial orbit under whose size is strictly less than . But from the fact that the size of any orbit must divide the size of the group, we have a nontrivial divisor of the order of that is strictly smaller than , contradicting the least prime divisor assumption on .